Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A belt runs on a 160 mm driving pulley at 720 rpm. The tension is 1500 N on the tight side and 750 N on the slack side. Find the power P transmitted, in kW.
d1 = 160 mmn1 = 720 rpmT1 = 1500 NT2 = 750 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 120 mm pulley and a 450 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 120 mmd2 = 450 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 200 N·md = 30 mmL = 1.2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 50 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.
P = 50 kWN = 500 rpmdo = 75 mmdi = 56 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = 90 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −10 MPaσy = 90 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 110 MPa, σᵧ = 110 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.
σx = 110 MPaσy = 110 MPaτxy = 20 MPaθ = 40°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa). Both are 500 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø25 mm, E = 200 GPatube: Ø120/Ø50 mm, E = 100 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 30 kNsegment 1: Ø35 mm, 1 m, brass (E = 100 GPa)segment 2: Ø16 mm, 2 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 95 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 95 kNn = 3L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 2 cycles later by a peak of 4.4 mm. Find the damping ratio ζ.
x0 = 12 mmx2 = 4.4 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 10 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 100 kN/mk2 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 4.524 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt speed
v = 6.032 m/s
Power
P = (1500 − 750)·6.032 = 4524 W = 4.524 kW
Answer
P = 4.524 kW
2.Gears & gear trains
Answer: θ = 141.5 °
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt angle
asin(0.33) = 19.27°
Angle of wrap
θ = 180 − 2·19.27 = 141.5°
Answer
θ = 141.5°
3.Torsion of shafts
Answer: φ = 2.162 °
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Polar second moment of area
J = π·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·1200/(80000·79522) = 0.03773 rad
φ = 0.03773·180/π = 2.162°
Answer
φ = 2.162°
4.Torsion of shafts
Answer: τmax = 16.73 MPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Torque
ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 50000/52.36 = 954.9 N·m
Polar second moment of area
J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·37.5/2.141 × 10⁶ = 16.73 MPa
Answer
τmax = 16.73 MPa
5.Mohr’s circle & stress transformation
Answer: θs = 70.1 °
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −60/78.1 = −0.7682, sin 2θs = −(−50)/78.1 = 0.6402
2θs = atan2(50, −60) = 140.2°
Angle on the element
θs = 70.1°
check: θs − θp = 70.1 − (−64.9) = 135°
Answer
θs = 70.1°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 129.7 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Transformation equation
σθ = 110 + 0·0.1736 + 20·0.9848 = 129.7 MPa
Answer
σθ = 129.7 MPa
7.Axial stress & strain
Answer: δ = 0.04841 mm
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
Ar = π·25²/4 = 490.9 mm²
At = π(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 100·10³·9346 = 1.033 × 10⁹ N
Shortening
δ = 100000·500/1.033 × 10⁹ = 0.04841 mm
Answer
δ = 0.04841 mm
8.Axial stress & strain
Answer: δ = 1.804 mm
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 31.18 MPa
σ2 = 149.2 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 31.18·1000/100000 = 0.3118 mm
δ2 = σ2L2/E2 = 149.2·2000/200000 = 1.492 mm
δ = δ1 + δ2 = 1.804 mm
Answer
δ = 1.804 mm
9.Column buckling
Answer: Pcr = 992.2 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/2000² = 992201 N = 992.2 kN
Answer
Pcr = 992.2 kN
10.Column buckling
Answer: d = 120.4 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Required critical load
Pcr = n·P = 3·95000 = 285000 N
KL = 1·5000 = 5000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·285000·5000²/(π³·70000))^(1/4) = 120.4 mm
Answer
d = 120.4 mm
Check: the slenderness KL/r is 166.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.07959
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Logarithmic decrement
δ = (1/2)·ln(12/4.4) = 0.5017
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5017/√(39.48 + 0.2517) = 0.07959
Answer
ζ = 0.07959
12.Mechanical vibrations
Answer: fn = 2.146 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/100 + 1/10 → keq = 100·10/(100 + 10) = 9.091 kN/m
Natural frequency
ωn = √(9091/50) = 13.48 rad/s
fn = 13.48/(2π) = 2.146 Hz
Answer
fn = 2.146 Hz
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